Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $19$ years; the standard deviation is $1.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living between $17.5$ and $23.5$ years.
Solution: $19$ $17.5$ $20.5$ $16$ $22$ $14.5$ $23.5$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $19$ years. We know the standard deviation is $1.5$ years, so one standard deviation below the mean is $17.5$ years and one standard deviation above the mean is $20.5$ years. Two standard deviations below the mean is $16$ years and two standard deviations above the mean is $22$ years. Three standard deviations below the mean is $14.5$ years and three standard deviations above the mean is $23.5$ years. We are interested in the probability of a gorilla living between $17.5$ and $23.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the gorillas will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the mean. The probability of a particular gorilla living between $17.5$ and $23.5$ years is ${68\%} + \color{orange}{15.85\%}$, or $83.85\%$.